Question

A ground state hydrogen atom absorbs a photon light having a wavelength of 92.57 nm. It then gives off a photon having a wavelength of 1944 nm. What is the final state of the hydrogen atom?

I got Nf=1.14 and it wasn't right

Answer 1

Absorbed photon energy Ea = hc/λ.. (Planck's equation) =Ea = hc / 92.57x10^-9m

Energy emitted .. Ee = hc/ 1944x10^-9m

Energy retained .. ∆E = Ea - Ee =hc(1/92.57^-9 - 1/1944^-9)
∆E = hc(10⁹/92.57-10⁹/1944)

=hcx10⁹(0.0108-0.01.286)

∆E =(6.625x10^-34)(3x10⁸)x10⁷x1.0286

∆E =20.40x10^-19

∆E =2.040x10^-18J

converting J to eV

∆E =2.040x10^-18/1.60x10^-19

∆E =12.75eV

New energy state = (-13.60 + 12.75)eV = -0.85 eV

Energy states for Hydrogen .. En = - (13.60 / n²)

n² = -13.60 / -0.85 = 16 .. .. .. ►n = 4